Define a positive integer $n$ to be a factorial tail if there is some positive integer $m$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?

Explanation: Let the number of zeros at the end of $m!$ be $f(m)$. We have $f(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \left\lfloor \frac{m}{125} \right\rfloor + \left\lfloor \frac{m}{625} \right\rfloor + \left\lfloor \frac{m}{3125} \right\rfloor + \cdots$.
Note that if $m$ is a multiple of $5$, $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$.
Since $f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots  = \frac{m}{4}$, a value of $m$ such that $f(m) = 1991$ is greater than $7964$. Testing values greater than this yields $f(7975)=1991$.
There are $\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$, less than $1992$. Thus, there are $1991-1595 = \boxed{396}$ positive integers less than $1992$ that are not factorial tails.